\section{\texorpdfstring{关于求解$CB\Delta r$的推论}{关于求解CBΔr的推论}}
\subsection{求直接量的情况}
\subsubsection{求C.}
\begin{itemize}
  \item $C=B+\Delta$
  \item $C=B(1+r)$
  \item $C= \frac{\Delta}{r}(1+r) = \frac{\Delta}{r}+\Delta$
\end{itemize}

\subsubsection{求B.}
\begin{itemize}
  \item $B=C-\Delta$
  \item $B=\frac{C}{1+r}$
  \item $B=\frac{\Delta}{r}$
\end{itemize}

\subsubsection{\texorpdfstring{求$\Delta$.}{求Δ}}
\begin{itemize}
  \item $\Delta=C-B$
  \item $\Delta=\frac{Cr}{1+r}$
  \item $\Delta=Br$
\end{itemize}

\subsubsection{求r.}
\begin{itemize}
  \item $r=\frac{C-B}{B} = \frac{C}{B}-1$
  \item $r=\frac{\Delta}{C-\Delta}$
  \item $r=\frac{\Delta}{B}$
\end{itemize}

\subsection{\texorpdfstring{$B=B_1+B_2$的和差情况}{B=B1+B2的和差情况}}
\subsubsection{\texorpdfstring{求$C=C_1+C_2$.}{求C=C1+C2.}}
\begin{itemize}
  \item $C=(B_1+\Delta_1)+(B_2+\Delta_2)$
  \item $C=B_1(1+r_1)+B_2(1+r_2)$
  \item \begin{align}
          C
           & =\frac{\Delta_1}{r_1}(1+r_1)+\frac{\Delta_2}{r_2}(1+r_2)       \\
           & =\frac{\Delta_1}{r_1}+\Delta_1 + \frac{\Delta_2}{r_2}+\Delta_2 \\
           & =\frac{r_2\Delta_1+r_1\Delta_2}{r_1r_2}+\Delta_1+\Delta_2
        \end{align}
\end{itemize}

\subsubsection{\texorpdfstring{求$B=B_1+B_2$.}{求B=B1+B2.}}
\begin{itemize}
  \item $B=(C_1-\Delta_1)+(C_2-\Delta_2)$
  \item $B=\frac{C_1}{1+r_1}+\frac{C_2}{1+r_2}$
  \item $B=\frac{\Delta_1}{r_1} + \frac{\Delta_2}{r_2}$
\end{itemize}

\subsubsection{\texorpdfstring{求$\Delta=\Delta_1+\Delta_2$.}{求Δ=Δ1+Δ2}}
\begin{itemize}
  \item $\Delta=(C_1+C_2)-(B_1+B_2)$
  \item $\Delta=(C_1+C_2)-(\frac{C_1}{1+r_1}+\frac{C_2}{1+r_2})=\frac{C_1r_1}{1+r_1}+\frac{C_2r_2}{1+r_2}$
  \item $\Delta=B_1r_1+B_2r_2$
\end{itemize}

\subsubsection{求r.}
\begin{itemize}
  \item $r=\frac{C_1+C_2}{B_1+B_2}-1$
  \item $r=\frac{\Delta_1+\Delta_2}{(C_1+C_2)-(\Delta_1+\Delta_2)}$
  \item $r=\frac{\Delta_1+\Delta_2}{B_1+B_2}$
  \item $r=\frac{B_1r_1+B_2r_2}{B_1+B_2}$ (加权平均，$B_i$为$r_i$的权重，$B_i$越大，$r_i$对最终平均值的影响就越大。)
  \item \begin{align}
          r
           & =\frac{\Delta_1+\Delta_2}{\frac{\Delta1}{r_1}+\frac{\Delta_2}{r_2}} \\
           & =(\Delta_1+\Delta_2)\frac{r_1r_2}{r_2\Delta_1+r_1\Delta_2}
        \end{align}
\end{itemize}
\begin{lemma}\label{lem:r_analysis_weighted_average}
  和差型的总体增长率r，必然是两个部分增长率$r_1$和$r_2$的加权平均值。这意味着：
  \begin{enumerate}
    \item r的值必定介于$r_1$和$r_2$之间。
    \item r的值会更偏向于基期量更大的那个部分的增长率。
  \end{enumerate}
\end{lemma}
\begin{lemma}\label{lem:lever_principle}
  由杠杆原理可知$F_1\times L_1 = F_2\times L_2$，则$\frac{F_1}{F_2}=\frac{L_2}{L_1}$，即量与距离成反比。
\end{lemma}
\begin{theorem}
  由\cref{lem:r_analysis_weighted_average,lem:lever_principle}，假设$r_2>r_1$，可得$\frac{r-r_1}{r_2-r}=\frac{B_2}{B_1}$，那么$r=r_1+(r_2-r_1)\frac{B_2}{B_1+B_2}$，即
  $$最终值 = 左端点值+总距离\times \frac{右端点权重}{总权重}$$
\end{theorem}

\subsection{\texorpdfstring{$B=B_1 \times B_2$的乘积情况}{B=B1*B2的乘积情况}}
\subsubsection{\texorpdfstring{求$C=C_1 \times C_2$.}{求C=C1*C2}}
\begin{itemize}
  \item $C=(B_1+\Delta_1) \times (B_2+\Delta_2)$
  \item $C=B_1(1+r_1) \times B_2(1+r_2) $
  \item $C= \frac{\Delta_1}{r_1}(1+r_1) \times \frac{\Delta_2}{r_2}(1+r_2)$
\end{itemize}

\subsubsection{\texorpdfstring{求$B=B_1 \times B_2$.}{求B=B1*B2}}
\begin{itemize}
  \item $B=(C_1-\Delta_1) \times (C_2-\Delta_2)$
  \item $B=\frac{C_1}{1+r_1} \times \frac{C_2}{1+r_2}$
  \item $B=\frac{\Delta_1}{r_1} \times \frac{\Delta_2}{r_2}$
\end{itemize}

\subsubsection{\texorpdfstring{求$\Delta$.}{求Δ=Δ1*Δ2}}
\begin{itemize}
  \item $\Delta=(C_1C_2)-(B_1B_2)$
  \item $\Delta=(C_1C_2)-(\frac{C_1}{1+r_1}\frac{C_2}{1+r_2})$
  \item $\Delta=(B_1B_2)(r_1+r_2+r_1r_2)$
  \item $\Delta=B_1\Delta_2 + B_2\Delta_1 + \Delta_1\Delta_2$
  \item $\Delta=\Delta_1\Delta_2(\frac{1}{r_1}+\frac{1}{r_2}+1)$
  \item $\Delta=C_1\Delta_2+C_2\Delta_1-\Delta_1\Delta_2$
\end{itemize}

\subsubsection{求r(乘积增长率)：\texorpdfstring{$r=r_1+r_2+r_1r_2$}{r=r1+r2+r1r2}.}
\begin{itemize}
  \item $r=\frac{C_1C_2}{B_1B_2}-1$
  \item $r=\frac{C_1\Delta_2+C_2\Delta_1-\Delta_1\Delta_2}{(C_1-\Delta_1) \times (C_2-\Delta_2)}$
  \item $r=\frac{B_1\Delta_2 + B_2\Delta_1 + \Delta_1\Delta_2}{B_1B_2}$
  \item $r=\frac{(B_1B_2)(r_1+r_2+r_1r_2)}{B_1B_2}=r_1+r_2+r_1r_2$
\end{itemize}

\subsection{\texorpdfstring{$B=\frac{B_1}{B_2}$}{B=B1/B2}的比值情况}
\subsubsection{求\texorpdfstring{$C=\frac{C_1}{C_2}$}{C=C1/C2}.}
\begin{itemize}
  \item $C=\frac{B_1(1+r_1)}{B_2(1+r_2)}$
  \item $C=\frac{B_1+\Delta_1}{B_2+\Delta_2}$
  \item $C=\frac{\Delta_1(1+\frac{1}{r_1})}{\Delta_2(1+\frac{1}{r_2})} = \frac{\Delta_1}{\Delta_2}\frac{1+r_1}{r_1}\frac{r_2}{1+r_2}$
\end{itemize}

\subsubsection{求B(基期比值、比重): \texorpdfstring{$B=\frac{B_1}{B_2}$}{B=B1/B2}.}
\begin{itemize}
  \item $B=\frac{C_1}{C_2}\frac{1+r_2}{1+r_1}$
  \item $B=\frac{C_1-\Delta_1}{C_2-\Delta_2}$
  \item $B=\frac{\Delta_1}{\Delta_2}\frac{r_2}{r_1}$
\end{itemize}

\subsubsection{求\texorpdfstring{$\Delta$}{Δ}(比值差、比重差).}
\begin{itemize}
  \item $\Delta=\frac{B_1}{B_2}\frac{r_1-r_2}{1+r_2}=\frac{B_1}{C_2}(r_1-r_2)$
  \item $\Delta=\frac{C_1}{C_2}\frac{r_1-r_2}{1+r_1}$
  \item $\Delta=\frac{\Delta_1}{\Delta_2}\frac{r_2}{r_1}\frac{r_1-r_2}{1+r_2}$
  \item $\Delta=\frac{B_1+\Delta_1}{B_2+\Delta_2}-\frac{B_1}{B_2} =\frac{B_2\Delta_1-B_1\Delta_2}{B_2(B_2+\Delta_2)}$
  \item $\Delta=\frac{C_1}{C_2}-\frac{C_1-\Delta_1}{C_2-\Delta_2}=\frac{C_2\Delta_1-C_1\Delta_2}{C_2(C_2-\Delta_2)}$
\end{itemize}

\subsubsection{求r(比值增长率): \texorpdfstring{$r=\frac{r_1-r_2}{1+r_2}$}{r=(r1-r2)/(1+r2)}.}
\begin{itemize}
  \item $r=\frac{\frac{B_1(1+r_1)}{B_2(1+r_2)}}{\frac{B_1}{B_2}}-1=\frac{1+r_1}{1+r_2}-1=\frac{r_1-r_2}{1+r_2}$
  \item $r=\frac{\frac{C_1}{C_2}}{\frac{C_1-\Delta_1}{C_2-\Delta_2}}-1$
  \item $r=\frac{\frac{B_1+\Delta_1}{B_2+\Delta_2}}{\frac{B_1}{B_2}}-1$
  \item $r=\frac{\frac{C_1}{C_2}}{\frac{B_1}{B_2}}-1$
\end{itemize}

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